See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the reaction: 1-methylcyclohex-1-ene reacts with Br2 in H2O. This is a bromohydrin formation reaction. In the presence of water, Br2 forms a bromonium ion intermediate, and water acts as the nucleophile instead of bromide. Step 2 – Bromonium ion formation: Br2 attacks the double bond of 1-methylcyclohex-1-ene to form a cyclic bromonium ion. The bromonium ion bridges the two carbons of the former double bond (C1 and C2). Because C1 is more substituted (it bears the methyl group), the bromonium ion is unsymmetrical and the positive charge is more developed at C1. Step 3 – Regioselectivity of nucleophilic attack: Water (the nucleophile) attacks the more electrophilic, more substituted carbon (C1, which bears the CH3 group) in an anti (backside) fashion relative to the bromonium ion. This gives the OH at C1 on the face opposite to the bromine bridge. Meanwhile, Br ends up at C2. Step 4 – Stereochemistry: Anti addition means OH and Br are added to opposite faces of the original double bond. Water attacks C1 from the face opposite the bromonium (anti), placing OH on one face at C1. Bromine is then delivered to C2 from the bromonium on the opposite face. This gives a trans relationship between the OH (at C1) and Br (at C2) — i.e., they are on opposite faces. Step 5 – Product structure: The major product has: - C1: CH3 (wedge) and OH (wedge) — both on same face because water attacked C1 anti to the bromine bridge, placing OH on the same side as CH3 if CH3 is already on that face, resulting in CH3 and OH both on wedge at C1. - C2: Br at the bottom (same side as the bromonium was, i.e., opposite to OH) and H on dash. This corresponds to option (d): cyclohexane ring with CH3 on wedge at C1, OH on wedge at C1, H on dash at C2, Br at bottom of C2. Step 6 – Why other options fail: - (a) and (c) place Br at C1, which would require syn addition or rearrangement — not the bromohydrin mechanism. - (b) places Br on dash at C2 and OH on wedge at C1, but the relative orientation of H at the bottom and Br on dash at C2 gives a different stereochemical relationship than what anti addition across the bromonium predicts for the major product. - (d) correctly shows anti addition with OH at the more substituted C1 (Markovnikov-like water attack) and Br at C2 in a trans relationship. Therefore, the correct answer is D.